# Catching and batting

When you catch a ball of mass m that was travelling at velocity v the change of momentum of the ball during the catching is mv. To do this you must apply a force F for a time t. So since impulse equals change of momentum Ft = mv. You can see that the longer you take to bring the ball to rest the smaller the force - that's why it is sensible to "bring your hands back" when catching a ball - especially a hard one like a cricket ball, the ball does not "give" so you must.

When you hit the ball straight back the change of momentum is much bigger.

Momentum of ball before collision = mv
Momentum of ball after collision = - mv [momentum is a vector and we must allow for the change of direction, the minus sign tells you that the ball is moving in the opposite direction].

Therefore change of momentum of ball = mv - (-mv) = 2mv

So force on the ball F = 2mv/t and if the time spent in hitting the ball backwards is the same as that spent in catching you can see that the force will be twice as large.

Catching :-      Momentum change = mv      Force = mv/t

Hitting back at the same speed :-      Momentum change = 2mv      Force = 2mv/t

Example problems
1. A girl plays in a hockey match. At one point she stops a ball from a corner with her stick and at another she hits a ball that was coming towards her straight back. In both cases assume that the mass of the hockey ball was 0.15 kg, the velocity of the ball was 12 ms-1 and that it was in contact with her stick for 0.05 s. Calculate the force in each case:

(a) Stopping the ball: F = mv/t = 0.15x12/0.05 = 36 N
(b) Hitting the ball back: F = 2mv/t = 2x0.15x12/0.05 = 72 N

2. Emission of an alpha particle
A radium nucleus with a relative atomic mass of 226 (226Ra) emits an alpha particle (relative atomic mass 4 (4He)) at 1.5x107 ms-1 .
What is the recoil velocity of the radon nucleus?

Using: recoil velocity (V) = -vm/M = - 1.5x107x4/226 = - 2.65x105 ms-1

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